# Direction of average acceleration in circular motion

I know that the instantaneous acceleration is always directed towards the center of the circle.But what about average acceleration.

In the above figure my book says place change in velocity along the line that bisects angle $$r$$ and $$r'$$ and observe that it is directed towards centre.

my question is that is there any rule that we should place it along the angle bisector between the two given points to get average acceleration direction.

Any help will be appreciated

• Symmetry arguments Feb 16 '20 at 8:14
• The average acceleration is $\overline{a}=\dfrac{1}{t_{2}-t_{1}}\int ^{t_{2}}_{t_{1}}\dfrac{dv}{d\tau }d\tau =\dfrac{\Delta v}{\Delta t}=\dfrac{\Delta v}{\Delta \theta } \dfrac{\Delta \theta }{\Delta t}=\dfrac{\Delta V}{\Delta \theta } \dfrac{\Delta v}{\Delta R}$
– Eli
Feb 16 '20 at 9:07
• @Eli What are $V$ and $R$? Feb 16 '20 at 17:21
• Is figure (a) meant to represent motion that is not circular, and also not constant speed? The blue trace seems non-circular. Are we to assume that the magnitudes of ${\bf v}$ and ${\bf v'}$ are the same? Feb 16 '20 at 17:26
• It is actually easy and can be thought of qualitatively without getting involved into rigorous mathematics. Feb 16 '20 at 17:29

The average acceleration over one or more complete revolutions is zero. This can be seen from the fact that when the object is on the opposite side its acceleration is opposite.

If the average acceleration were nonzero, the object would not keep returning to the same location.

• I think the OP is asking about the direction of average acceleration over points not located diametrically opposite to each other. As he has shown in the fig. Feb 17 '20 at 4:38
• Perhaps so. Only the OP knows for sure, and the question seems vague to me. The average acceleration over any fraction of a circle can certainly be calculated, but I don’t see why it would be of any interest. Feb 17 '20 at 5:30

Acceleration is the derivative with respect to time of velocity. From definition $$\vec a(t) = \lim_{\Delta t \rightarrow 0} \frac{\vec v(t+\Delta t)- \vec v(t)}{\Delta t}.$$ The average acceleration between time $$t_1$$ and $$t_2$$ is defined as $$\vec a_m = \frac{1}{ (t_2-t_1)}\int_{t_1}^{t_2} \vec a(t) \mathrm{d}t.~~~~~~~[1]$$ Recalling that velocity is the primitive function of acceleration, this definition is equivalent to $$\vec a_m = \frac{\vec v(t_2)- \vec v(t_1)}{ t_2-t_1}.~~~~~~~~~~[2]$$ It is clear that the average acceleration $$a_m$$ is a property of the interval $$[t_1,t_2]$$. Therefore, one could ask the question: is there a time $$\tau$$, in the interval $$[t_1,t_2]$$, such that the average acceleration $$\vec a_m$$ is the best possible approximation of $$\vec a(\tau)$$? Lagrange theorem provides a strong, affirmative answer. Under reasonable hypotheses of regularity, there is a time $$\tau$$, in the interval $$[t_1,t_2]$$, such that $$\vec a(\tau) = \vec a_m$$. Unfortunately, in general it is not easy to find such a point without full knowledge of the function $$\vec a(t)$$ and solving an implicit equation fo $$\tau$$.

However, in general, it is possible to get a uniform, optimized approximation by choosing $$\tau = (t_1 + t_2)/2$$. It is possible to show that with such a choice (corresponding to the symmetric difference formula for approximating numerically the first derivative of a function), one is nullifying the first order error in $$\Delta t$$, leaving only an error $$O(\Delta t^2)$$.

These considerations hold for every possible motion. In the special case of a uniform circular motion, symmetry in time of the two-point formula implies symmetry with respect to the angle (i.e. the bisector choice). Moreover, it turns out that the direction of the approximated acceleration is exact. The reason is is clear by going back to [$$1$$] and [$$2$$], in the special case of uniform circular motion. On the one hand, [$$1$$] becomes $$\vec a_m = \frac{1}{ 2 \Delta t}\int_{t - \Delta t}^{t+\Delta t} \vec a(t^{\prime})\mathrm{d}t^{\prime},$$ which, by symmetry, for all $$\Delta t < T/4$$, has the same direction as $$\vec a(t)$$, $$T$$ being the period of the motion. On the other hand, for the same interval of $$\Delta t$$, $$\vec v(t+\Delta t)- \vec v(t-\Delta t)$$, again for symmetry, has to point in the direction of $$\vec a(t)$$.

That should explain the reason for the choice.

Since the average acceleration is along Δv (a=Δv/Δt), the average acceleration is perpendicular to Δr. We already know that, since the path is circular, v is perpendicular to r and so is v' to r', according to the figure given by you. (Since the velocity vectors v and v' are always perpendicular to the position vectors, the angle between them is also ΔΘ).

Note that the book stated that If we place Δv on the line that bisects the angle between r and r', we see that it is directed towards the centre of the circle. It was just a verification of the statement already quoted above.

A perpendicular from the centre bisects the chord, and since PCP' is an isosceles triangle, bisects the angle between r and r' (from the angle bisector theorem). Obviously it will be directed towards the centre considering the geometry of the figure,

or understand it this way,

The perpendicular bisector of a chord passes through the centre of the circle. It is a very fundamental theorem which has many applications in various fields of Physics and Mathematics. I have attached a link to its proof .

Actually, it is not a rule, it is just an approach to introduce this topic at the beginner level, assuming that you know basic geometry, upto 10th grade.

In a nutshell, since Δv is perpendicular to Δr, it passes through the centre of the circle.

And so, the average acceleration is directed towards the centre.

Note that the average acceleration only changes into instantaneous acceleration if we put the limits (Δt->0). So, its direction is towards the centre. In the fig(c), we are only approximating the situation on an infinitesimally small scale. But the overall concept remains the same.

• if Δv is perpendicular to Δr does it necessarily means that it is also perpendicular bisector of Δr. Feb 17 '20 at 2:45
• @utkarshbhatt Yes. In this case. Read my answer carefully. Feb 17 '20 at 4:35
• It naturally has to pass through the centre. Feb 17 '20 at 4:36

the equation for average acceleration is: $$\vec{a}_A=\frac{1}{t_2-t_1}\int_{t_1}^{t_2}\,\frac{d\vec{v}}{dt}\,dt= \frac{1}{t_2-t_1}\int_{t_1}^{t_2}\,d\vec{v}=\frac{\vec{v}(t_2)-\vec{v}(t_1)}{t_2-t_1}= \frac{\Delta \vec{v}}{\Delta t}$$

with the position vector :

$$\vec{R}(\tau)=\left[ \begin {array}{c} r \left( \tau \right) \cos \left( \varphi \left( \tau \right) \right) \\ r \left( \tau \right) \sin \left( \varphi \left( \tau \right) \right) \end {array} \right]$$ and the velocity $$\vec{v}=\frac{d\vec{R}}{d\tau}$$:

$$\vec{v}= \left[ \begin {array}{c} \cos \left( \varphi \left( \tau \right) \right) {\frac {d}{d\tau}}r \left( \tau \right) -r \left( \tau \right) \sin \left( \varphi \left( \tau \right) \right) {\frac {d}{ d\tau}}\varphi \left( \tau \right) \\ \sin \left( \varphi \left( \tau \right) \right) {\frac {d}{d\tau}}r \left( \tau \right) +r \left( \tau \right) \cos \left( \varphi \left( \tau \right) \right) {\frac {d}{d\tau}}\varphi \left( \tau \right) \end {array} \right]$$

thus:

$$\Delta\vec{v}=\lim_{\tau=t_2}\vec{v}(\tau)- \lim_{\tau=t_1}\vec{v}(\tau)$$

for a circle $$r(\tau)=\rho\quad$$ and $$\varphi(\tau)=a\,\tau$$

you get $$\Delta\vec{v} =\left[ \begin {array}{c} -\rho\,\sin \left( at_{{2}} \right) a+\rho\, \sin \left( at_{{1}} \right) a\\ \rho\,\cos \left( a t_{{2}} \right) a-\rho\,\cos \left( at_{{1}} \right) a\end {array} \right]$$

so if $$t_2=\frac{3\pi}{a}\quad,t_1=\frac{\pi}{a}$$ thus $$\Delta\vec{v}=\vec{0}$$ and the average acceleration is zero.

but for example $$\varphi(\tau)=a\,\tau^2$$ you can't obtain the values of $$t_2\,,t_1$$ to get the average acceleration be zero.